A user of RSA creates and then publishes a public key based on the two large prime numbers, along with an auxiliary value. The prime numbers must be kept secret. Anyone can use the public key to encrypt a message, but with currently published methods, if the public key is large enough, only someone with knowledge of the prime factors can feasibly decode the message. Breaking RSA encryption is known as the RSA problem. It is an open question whether it is as hard as the factoring problem.

Key generation

RSA involves a public key and a private key. The public key can be known by everyone and is used for encrypting messages. Messages encrypted with the public key can only be decrypted in a reasonable amount of time using the private key. The keys for the RSA algorithm are generated the following way:

e having a short bit-length and small Hamming weight results in more efficient encryption – most commonly 216 + 1 = 65,537. However, much smaller values of e (such as 3) have been shown to be less secure in some settings.[5]

The public key consists of the modulus n and the public (or encryption) exponent e. The private key consists of the modulus n and the private (or decryption) exponent d, which must be kept secret. p, q, and φ(n) must also be kept secret because they can be used to calculated.

An alternative, used by PKCS#1, is to choose d matching de ≡ 1 (mod λ) with λ = lcm(p − 1, q − 1), where lcm is the least common multiple. Using λ instead of φ(n) allows more choices for d. λ can also be defined using the Carmichael function, λ(n).

The public key is (n = 3233, e = 17). For a padded plaintext message m, the encryption function is

The private key is (n = 3233, d = 2753). For an encrypted ciphertextc, the decryption function is

For instance, in order to encrypt m = 65, we calculate

To decrypt c = 2790, we calculate

Both of these calculations can be computed efficiently using the square-and-multiply algorithm for modular exponentiation. In real-life situations the primes selected would be much larger; in our example it would be trivial to factor n, 3233 (obtained from the freely available public key) back to the primes p and q. Given e, also from the public key, we could then compute d and so acquire the private key.

Practical implementations use the Chinese remainder theorem to speed up the calculation using modulus of factors (mod pq using mod p and mod q).

The values dp, dq and qinv, which are part of the private key are computed as follows:

Here is how dp, dq and qinv are used for efficient decryption. (Encryption is efficient by choice of public exponent e)

Signing messages

Suppose Alice uses Bob‘s public key to send him an encrypted message. In the message, she can claim to be Alice but Bob has no way of verifying that the message was actually from Alice since anyone can use Bob’s public key to send him encrypted messages. In order to verify the origin of a message, RSA can also be used to sign a message.

Suppose Alice wishes to send a signed message to Bob. She can use her own private key to do so. She produces a hash value of the message, raises it to the power of d (modulo n) (as she does when decrypting a message), and attaches it as a “signature” to the message. When Bob receives the signed message, he uses the same hash algorithm in conjunction with Alice’s public key. He raises the signature to the power of e (modulo n) (as he does when encrypting a message), and compares the resulting hash value with the message’s actual hash value. If the two agree, he knows that the author of the message was in possession of Alice’s private key, and that the message has not been tampered with since.

Proof using Fermat’s little theorem

The proof of the correctness of RSA is based on Fermat’s little theorem. This theorem states that if p is prime and p does not divide an integer a then

We want to show that (me)d ≡ m (mod pq) for every integer m when p and q are distinct prime numbers and e and d are positive integers satisfying

We can write

for some nonnegative integer h.

To check two numbers, like med and m, are congruent mod pq it suffices (and in fact is equivalent) to check they are congruent mod p and mod q separately. (This is part of the Chinese remainder theorem, although it is not the significant part of that theorem.) To showmed ≡ m (mod p), we consider two cases: m ≡ 0 (mod p) and m 0 (mod p).

In the first case med is a multiple of p, so med ≡ 0 ≡ m (mod p). In the second case

The verification that med ≡ m (mod q) proceeds in a similar way, treating separately the cases m ≡ 0 (mod q) and m 0 (mod q), using Fermat’s little theorem for modulus q in the second case.

This completes the proof that, for any integer m,

Proof using Euler’s theorem

Although the original paper of Rivest, Shamir, and Adleman used Fermat’s little theorem to explain why RSA works, it is common to find proofs that rely instead on Euler’s theorem.

We want to show that med ≡ m (mod n), where n = pq is a product of two different prime numbers and e and d are positive integers satisfying ed ≡ 1 (mod φ(n)). Since e and d are positive, we can write ed = 1 + hφ(n) for some non-negative integer h. Assuming that m is relatively prime to n, we have

When m is not relatively prime to n, the argument just given is invalid. This is highly improbable (only a proportion of 1/p + 1/q − 1/(pq) numbers have this property), but even in this case the desired congruence is still true. Either m ≡ 0 (mod p) or m ≡ 0 (mod q), and these cases can be treated using the previous proof.